Let $d > n$$d < n$, and suppose that $G \in \mathbb{R}^{n \times d}$ is a random matrix with iid standard Normal entries. Letlet $K_G$$G_n(d)$ denote the (random) kernelspace of this matrix. It has codimensionall $n$ with probability$d$-dimensional subspaces of $1$$\mathbb{R}^n$.
Consider the following ``eigenvalue withinLet $a = (a_1,\dots, a_n)$ denote a subspace'' for the diagonal operator corresponding to the positive sequence $a = (a_1, \dots, a_d)$, and define $$ \lambda_a(G) = \sup_{u \in K_G} \Big\{\, \sum_{j=1}^d u_j^2 : \sum_{j=1}^d \frac{u_j^2}{a_j} \leq 1\,\Big\} $$$U(a) = \{u \in \mathbb{R}^n: \sum_{j} u_j^2/a_j \leq 1\}$. Then define $$ \lambda_a(K) = \sup_{u \in K \cap U(a)} \sum_{j=1}^n u_j^2 $$
WhatI am interested in $\mu_a = \mathbb{E}[\lambda_a(K)]$, where the expectation is with respect to $\mathbb{E}[\lambda_a(G)]$$K$ drawn from the uniform (Haar) measure on $G_n(d)$. Is this possible to compute?
Two obvious observations:
- In the special case that $a_1 = a_2 = \cdots= a_d \equiv \alpha$, then evidently $$ \lambda_a(G) = \alpha, $$$$ \lambda_a(K) = \alpha, $$ with probability $1$, and the. The supremum is attained by taking any unit vector in $K_G$$K$ and scaling it appropriately.
- We can define $\tilde \lambda_a(K) = \sup_{u \in K} \{ \sum_{j=1}^d u_j^2 : \sum_{j=1}^d u_j^2/a_j \leq 1\}$In general, where $K \subset \mathbb{R}^d$ is a subspace of codimension $n$. Then $\mathbb{E}[\lambda_a(G)] = \mathbb{E}[\tilde \lambda_a(K)]$ where the latter expectation is with respect to Haar measure over $K$ with codimension$\lambda_a(K) \leq \max_j a_j$, and hence $n$$\mu_a \leq \max_j a_j$.
I have included the tag random-matrices as this problem can equivalently be formulated in terms of the kernel of a random matrix $G$ filled with standard Normal entries.